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- Question 17 Exercise 4.2 @math-11-kpk:sol:unit04
- KPTBB) Peshawar, Pakistan. =====Question 17===== There are $n$ arithmetic means between 5 and 32 such th
- Question 8 Exercise 4.2 @math-11-kpk:sol:unit04
- dfrac{a+b-c}{c}$ are in A.P, thus\\ \begin{align}\therefore \dfrac{c+a-b}{b}-\dfrac{b+c-a}{a}&=\dfrac{a+b
- Question 5 and 6 Exercise 4.2 @math-11-kpk:sol:unit04
- , in second term the power of $b$ is 1 and so on, therefore $$a_n=\log (a b^{n-1}).$$ We show that the g
- Question 5 and 6 Exercise 7.3 @math-11-kpk:sol:unit07
- ht) \\ & \times\left(1+\frac{5 x}{8}\right) \\ & \therefore\left(1-\frac{5 x}{4}\right)\left(1+\frac{5 x}
- Question 8 Exercise 7.2 @math-11-kpk:sol:unit07
- here } \\ & x^2=\frac{3}{2} \frac{3}{2}: 8 \\ & \therefore \frac{11}{1} \cdot i \quad 7 \\ & =5.82 \quad
- Question 6 Exercise 7.2 @math-11-kpk:sol:unit07
- cause $r$ should be a positive integer. It means there is no constant term or term independent of $x$ in
- Solutions: Math 11 KPK
- s in accordance with this plan (SLOs based). But there is no doubt that these solutions are valid for al
- Question 11 Review Exercise 6 @math-11-kpk:sol:unit06
- ccvers one fourth $\dfrac{1}{4}$ of the spinner. Therefore, \begin{align}P(\operatorname{Red})&=\dfrac{1... n} Also these two are mutually exclusive events. Therefore $P(R \cap G)=\phi$, where $R$ stands for red
- Question 9 & 10 Review Exercise 6 @math-11-kpk:sol:unit06
- In this case number should not start with $0$, therefore the total numbers that do not start with zero
- Question 9 Exercise 6.5 @math-11-kpk:sol:unit06
- n of one does not depend on the other sclection, therefore these two events are independent. Thus \begin... {5}\end{align} Only one is selected In this case there are two chance. it may be Ajmal selected Bushra
- Question 8 Exercise 6.5 @math-11-kpk:sol:unit06
- t the sum is $7$ or $11$ are mutually exclusive, therefore by addition law of probability we have \beg
- Question 7 Exercise 6.5 @math-11-kpk:sol:unit06
- {1}{13}$$ Since the events are mutually disjoints therefore, by addition law of probability we have \begi
- Question 5 and 6 Exercise 6.5 @math-11-kpk:sol:unit06
- ary events, and we are given $P(E)=\dfrac{8}{9}$, therefore, \begin{align}P(E^{\prime})&=1-P(E)=1-\dfrac{
- Question 3 and 4 Exercise 6.5 @math-11-kpk:sol:unit06
- that $\mathrm{A}$ and $B$ are mutually exclusive, therefore $A \cap B=\emptyset$. Thus \begin{align}P(A ... n} Álso \begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A \cap B)&=3\end{align} Now $$P(A)=\dfrac{
- Question 3 Exercise 6.4 @math-11-kpk:sol:unit06
- ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ay i.e. $${ }^8 C_8=\dfrac{8 !}{(8-8) ! 8 !}=1$$ Therefore probability to $8$ answers are correct is: $$... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc